Question: $\lim_{x\to\infty}\dfrac{\ln(x^2)}{(\ln(x))^2}=?$ Choose 1 answer: Choose 1 answer: (Choice A) A $0$ (Choice B) B $1$ (Choice C) C $4$ (Choice D) D $-\infty$
Answer: $\lim_{x\to\infty} \ln(x^2)=\infty$ and $\lim_{x\to\infty} (\ln(x))^2=\infty$, so $\lim_{x\to\infty}\dfrac{\ln(x^2)}{(\ln(x))^2}$ results in the indeterminate form $\dfrac{\infty}{\infty}$. We should use l'Hôpital's rule. $\begin{aligned} &\phantom{=}\lim_{x\to\infty}\dfrac{\ln(x^2)}{(\ln(x))^2} \\\\ &=\lim_{x\to\infty}\dfrac{\dfrac{d}{dx}\left[\ln(x^2)\right]}{\dfrac{d}{dx}[(\ln(x))^2]} \gray{\text{l'Hôpital's rule}} \\\\ &=\lim_{x\to\infty}\dfrac{\left(\dfrac{2x}{x^2}\right)}{2\ln(x)\cdot\dfrac{1}{x}} \\\\ &=\lim_{x\to\infty}\dfrac{1}{\ln(x)} \\\\ &=0 \end{aligned}$ Note that we were only able to use l'Hôpital's rule because the limit $\lim_{x\to\infty}\dfrac{\dfrac{d}{dx}\left[\ln(x^2)\right]}{\dfrac{d}{dx}[(\ln(x))^2]}$ can actually be determined. In conclusion, $\lim_{x\to\infty}\dfrac{\ln(x^2)}{(\ln(x))^2}=0$.